android sqlite查询判断是否有数据
答案:2 悬赏:10
解决时间 2021-04-28 17:08
- 提问者网友：騷女、無惡不作
- 2021-04-27 23:34
android sqlite查询判断是否有数据
最佳答案
- 二级知识专家网友：统治我的世界
- 2021-04-28 00:45
这个很简单的：
//打开或创建test.db数据库
SQLiteDatabasedb = openOrCreateDatabase("test.db", Context.MODE_PRIVATE, null);
//创建person表
db.execSQL("DROpTABLE IF EXISTS person");
db.execSQL("CREATE TABLE person (_idINTEGER PRIMARY KEY AUTOINCREMENT, name VARCHAR, age SMALLINT)");
//插入数据
Personperson = new Person();
person.name= "john";
person.age = 30;
db.execSQL("INSERT INTO person VALUES(NULL, ?, ?)",new Object[]{person.name, person.age});
//读取数据
Cursor c = db.rawQuery("SELECT* FROM person WHERe age >= ?", new String[]{"33"});
while (c.moveToNext()) {
int _id = c.getInt(c.getColumnIndex("_id"));
String name = c.getString(c.getColumnIndex("name"));
int age = c.getInt(c.getColumnIndex("age"));
Log.i("db", "_id=>" + _id + ", name=>" + name + ", age=>" + age);
}
c.close();
//关闭当前数据库
db.close();
//打开或创建test.db数据库
SQLiteDatabasedb = openOrCreateDatabase("test.db", Context.MODE_PRIVATE, null);
//创建person表
db.execSQL("DROpTABLE IF EXISTS person");
db.execSQL("CREATE TABLE person (_idINTEGER PRIMARY KEY AUTOINCREMENT, name VARCHAR, age SMALLINT)");
//插入数据
Personperson = new Person();
person.name= "john";
person.age = 30;
db.execSQL("INSERT INTO person VALUES(NULL, ?, ?)",new Object[]{person.name, person.age});
//读取数据
Cursor c = db.rawQuery("SELECT* FROM person WHERe age >= ?", new String[]{"33"});
while (c.moveToNext()) {
int _id = c.getInt(c.getColumnIndex("_id"));
String name = c.getString(c.getColumnIndex("name"));
int age = c.getInt(c.getColumnIndex("age"));
Log.i("db", "_id=>" + _id + ", name=>" + name + ", age=>" + age);
}
c.close();
//关闭当前数据库
db.close();
全部回答
- 1楼网友：情战凌云蔡小葵
- 2021-04-28 01:55
把它们放在一个adapter里面。
listadapter adapter = new simplecursoradapter(this,
android.r.layout.simple_list_item_2, c,
new string[] { phones.name, phones.number },
new int[] { android.r.id.text1, android.r.id.text2 });
setlistadapter(adapter);
你参考simplecursoradapter（）这个方法应该可以
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